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Thread: divisor

  1. #1
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    divisor

    Are we able to say, how do all the divisors of $\displaystyle x^m - 1 $ like???
    Thanks, nothing on my mind seems good now..
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  2. #2
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    Quote Originally Posted by sidi View Post
    Are we able to say, how do all the divisors of $\displaystyle x^m - 1 $ like???
    Thanks, nothing on my mind seems good now..
    yes. see cyclotomic polynomials.
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  3. #3
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    Thanks, but
    I havenīt heard about these polynomials before... I need it to prove that thing about d=gcd(m,n) $\displaystyle x^d - 1$ as the $\displaystyle gcd(x^m - 1,x^n - 1) $from yesterday. But I think I needn`t use these polynomials...there might be easier way to show it, but it can`t come on my mind...
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  4. #4
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    Quote Originally Posted by sidi View Post

    Thanks, but
    I havenīt heard about these polynomials before... I need it to prove that thing about d=gcd(m,n) $\displaystyle x^d - 1$ as the $\displaystyle gcd(x^m - 1,x^n - 1) $from yesterday. But I think I needn`t use these polynomials...there might be easier way to show it, but it can`t come on my mind...
    i see! the one that i gave you: $\displaystyle \gcd(x^n - 1, x^m -1)=x^{\gcd(n,m)} - 1.$ here's a proof: let $\displaystyle \gcd(n,m)=d.$ since $\displaystyle d \mid n$ and $\displaystyle d \mid m,$ we have: $\displaystyle x^d -1 \mid x^n - 1$ and $\displaystyle x^d - 1 \mid x^m - 1.$

    now suppose $\displaystyle f(x) \mid x^n - 1$ and $\displaystyle f(x) \mid x^m - 1.$ if we prove that $\displaystyle f(x) \mid x^d - 1,$ then we're done. we know that there exist $\displaystyle r,s \in \mathbb{N}$ such that $\displaystyle rn - sm=d.$ thus:

    $\displaystyle x^{rn} - 1 = x^d(x^m)^s - 1 \equiv x^d - 1 \mod f(x),$ because $\displaystyle x^m \equiv 1 \mod f(x).$ on the other hand $\displaystyle x^{rn}-1 \equiv 0 \mod f(x),$ because $\displaystyle f(x) \mid x^n - 1.$ hence $\displaystyle x^d - 1 \equiv 0 \mod f(x).$ Q.E.D.
    Last edited by NonCommAlg; Apr 14th 2009 at 04:04 PM.
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  5. #5
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    Quote Originally Posted by sidi View Post
    Thanks, but
    I havenīt heard about these polynomials before... I need it to prove that thing about d=gcd(m,n) $\displaystyle x^d - 1$ as the $\displaystyle gcd(x^m - 1,x^n - 1) $from yesterday. But I think I needn`t use these polynomials...there might be easier way to show it, but it can`t come on my mind...
    Here is a similar problem and an alternate solution.
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  6. #6
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    Thank you very much!
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